Maximum rotation angle of a rectangle inside a given rectangle I want to calculate the maximum rotation angle of a rectangle which is rotating with the center on the center of a bigger rectangle.  
Here is a figure for better understanding: fig.  
I have tried it out already but my solution is not plausible.
Thanks in advance!
Greets, Daniel
 A: Both rectangle have the same center due to their symmetry. There will be four contact points on the outer rectangle if the rectangles have same aspect ratio, else only two.
Let the inner rectangle have dimensions $(2a,2b)$. The y coordinate of contact point is $B_0/2$. Radius of inner rectangle is $ \sqrt{(a^2+b^2)} $ and its x- coordinate is $ \sqrt{(a^2+b^2)- (B_0/2)^2 } $
So coordinates of contact in first quadrant are 
$$  \sqrt{(a^2+b^2)- (B_0/2)^2 } ,\, B_0/2 $$
Now we can apply polar form of straight line to find $\alpha$
$$  x \cos \alpha + y \sin \alpha = b \tag1 $$
$$  \sqrt{(a^2+b^2)- (B_0/2)^2 }  \cos \alpha + (B_0/2)  \sin \alpha = b\tag2  $$
Let
$$ \beta= \tan ^{-1} \dfrac{(B_0/2)}{\sqrt{(a^2+b^2)- (B_0/2)^2 } }= \sin ^{-1} \dfrac{(B_0/2)}{\sqrt{(a^2+b^2) } }  \tag3 $$
Divide by $ \sqrt{a^2+b^2} $
$$ \cos(\alpha- \beta) = \frac{b}{\sqrt{a^2+b^2}} \tag4 $$
$$ \alpha -\beta = \cos^{-1}\frac{b}{\sqrt{a^2+b^2}}\tag5 $$
and finally find unknown rotation required,  $\theta = \alpha -\pi/2$
$$ \theta = \alpha - \pi/2 = \sin ^{-1} \dfrac{(B_0/2)}{\sqrt{(a^2+b^2) } } -  \sin^{-1}\frac{b}{\sqrt{a^2+b^2}}\tag6 $$
A: Find the diagonal lengt of the smaller rectangle, and find its projection onto a side of the larger rectangle.  
I did a lot of misguided calculation before hitting on the answer, but you should be able to do all the calculations in three simple steps.  
